Resource Development: Linear Programming and Scheduling for Edexcel A-Level Maths | 资源开发：Edexcel A-Level 数学中的线性规划与调度

📚 Resource Development: Linear Programming and Scheduling for Edexcel A-Level Maths | 资源开发：Edexcel A-Level 数学中的线性规划与调度

In Edexcel A-Level Mathematics, the phrase ‘resource development’ is best interpreted as the mathematical modelling of limited resources such as time, labour, materials and money. Although the term is not a standalone syllabus heading in pure mathematics, the core techniques appear in Decision Mathematics (D1) and in applied problem-solving questions. This article focuses on linear programming, resource histograms, resource leveling and critical path analysis – the key tools you need to model and optimise resource use under constraints.

在 Edexcel A-Level 数学中，“资源开发”最好理解为对时间、人力、材料和资金等有限资源的数学建模。虽然该术语不是纯数学考纲中的独立标题，但核心技巧出现在决策数学（D1）和应用问题中。本文重点讲解线性规划、资源直方图、资源均衡和关键路径分析——这些是你在约束条件下建模和优化资源使用所需的关键工具。

1. What Resource Development Means in A-Level Maths | A-Level 数学中资源开发的含义

In a mathematical context, resource development is not about mining or drilling. It is about deciding how to use scarce resources efficiently. You will usually be given constraints such as maximum machine hours, limited raw materials or a fixed number of workers, and asked to find the best possible outcome – maximum profit or minimum cost.

在数学语境中，资源开发不是指采矿或钻探，而是决定如何高效利用稀缺资源。你通常会得到诸如最大机器工时、有限原材料或固定工人数等约束条件，并被要求找出最佳结果——最大利润或最小成本。

Edexcel exam questions often present this as a linear programming problem, where a linear objective function is optimised subject to linear inequalities. In more advanced decision maths, resource development also covers scheduling and resource leveling for projects.

Edexcel 考试题目通常将其呈现为线性规划问题，即在线性不等式约束下优化线性目标函数。在更高级的决策数学中，资源开发还涵盖项目的调度和资源均衡。

2. Edexcel Syllabus Links | Edexcel 考纲对应

In Edexcel A-Level Further Maths, Decision Mathematics 1 covers the main techniques. Even if you are sitting A-Level Mathematics rather than Further Maths, linear programming questions can appear in applied contexts, so the skills are transferable.

在 Edexcel A-Level 进阶数学中，决策数学 1 涵盖了主要技巧。即使你参加的是 A-Level 数学而非进阶数学，线性规划问题也可能出现在应用情境中，因此这些技能是可迁移的。

3. Linear Programming: Formulating the Problem | 线性规划：问题建模

Start by defining your decision variables. Use x and y to represent the quantities of two resources or products, then write the objective function and every constraint as a linear expression.

首先定义决策变量。用 x 和 y 表示两种资源或产品的数量，然后将目标函数和每个约束写成线性表达式。

For example, if product A gives profit 3 per unit and product B gives profit 4 per unit, the objective is to maximise P = 3x + 4y.

例如，如果产品 A 每单位利润为 3，产品 B 每单位利润为 4，则目标是最大化 P = 3x + 4y。

Maximise P = 3x + 4y

Constraints often include machine time, material limits and demand. Always add non-negativity constraints x ≥ 0 and y ≥ 0 unless the question says otherwise.

约束通常包括机器时间、材料限制和需求量。除非题目另有说明，否则始终添加非负约束 x ≥ 0 和 y ≥ 0。

4. Graphing Inequalities and the Feasible Region | 图解不等式与可行域

To graph a constraint such as 2x + y ≤ 12, first draw the boundary line 2x + y = 12. Then test a point, usually (0,0), to decide which side of the line satisfies the inequality.

要绘制 2x + y ≤ 12 这样的约束，首先画出边界线 2x + y = 12。然后检验一个点（通常为 (0,0)），确定直线的哪一侧满足不等式。

The feasible region is the intersection of all shaded regions. If the problem is unbounded or empty, check your constraints again.

可行域是所有已着色区域的交集。如果问题无界或为空，请重新检查约束条件。

In Edexcel exams you must shade the region that is not satisfied, or clearly label the feasible region as R. Always use a ruler and write coordinates of key points on your diagram.

在 Edexcel 考试中，你必须对不满足的区域进行着色，或者将可行域清楚地标记为 R。务必使用直尺，并在图上写出关键点的坐标。

5. Vertex Testing and Optimal Solution | 顶点检验与最优解

The optimal solution of a linear programming problem occurs at a vertex of the feasible region. Calculate the coordinates of all vertices and substitute them into the objective function.

线性规划问题的最优解出现在可行域的顶点处。计算所有顶点的坐标，并将它们代入目标函数。

For example, suppose the feasible region has vertices (0,0), (6,0), (4,4) and (0,5). Evaluate P = 3x + 4y at each vertex:

例如，假设可行域的顶点为 (0,0)、(6,0)、(4,4) 和 (0,5)。在每个顶点处计算 P = 3x + 4y：